Question: The equation of a circle $C$ is $x^2+y^2+16x-4y+52 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+16x) + (y^2-4y) = -52$ $(x^2+16x+64) + (y^2-4y+4) = -52 + 64 + 4$ $(x+8)^{2} + (y-2)^{2} = 16 = 4^2$ Thus, $(h, k) = (-8, 2)$ and $r = 4$.